| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2007 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Generalised Binomial Theorem |
| Type | Factoring out constants before expansion |
| Difficulty | Standard +0.3 This is a standard C4 binomial expansion question with routine factoring of constants. Part (a) is direct application of the formula, part (b)(i) requires factoring out 8 and substituting, and part (b)(ii) involves a straightforward numerical substitution (x=1). All steps are textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.04c Extend binomial expansion: rational n, |x|<11.04d Binomial expansion validity: convergence conditions |
| Answer | Marks | Guidance |
|---|---|---|
| \((1+x)^{\frac{1}{3}} = 1 + \frac{1}{3}x + \frac{1}{3}\!\left(-\frac{2}{3}\right)\!\frac{1}{2}x^2\) | M1 | \(1 + \frac{1}{3}x + kx^2\) |
| \(= 1 + \frac{1}{3}x - \frac{1}{9}x^2\) | A1 (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sqrt[3]{8}\!\left(1 + \frac{3}{8}x\right)^{\frac{1}{3}}\) | B1 | \(8^{\frac{1}{3}}(1+kx)^{\frac{1}{3}}\) |
| \(= 2\!\left(1 + \frac{1}{3}\!\left(\frac{3}{8}x\right) - \frac{1}{9}\!\left(\frac{3}{8}x\right)^2\right)\) | M1 | Replacing \(x\) with \(kx\) in answer to (a) |
| \(= 2 + \frac{1}{4}x - \frac{1}{32}x^2\) | A1 (3 marks) | For numerical expression which would evaluate to answer given |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = \frac{1}{3}\): \(\sqrt[3]{8+1} = 2 + \frac{1}{4}\times\frac{1}{3} - \frac{1}{32}\times\left(\frac{1}{3}\right)^2\) | M1 | Using \(x = \frac{1}{3}\) in given answer |
| \(\sqrt[3]{9} = \frac{576+24-1}{288} = \frac{599}{288}\) | A1 (2 marks) | Any correct numerical expression \(= \frac{599}{288}\) |
## Question 5:
**Part (a)**
$(1+x)^{\frac{1}{3}} = 1 + \frac{1}{3}x + \frac{1}{3}\!\left(-\frac{2}{3}\right)\!\frac{1}{2}x^2$ | M1 | $1 + \frac{1}{3}x + kx^2$
$= 1 + \frac{1}{3}x - \frac{1}{9}x^2$ | A1 (2 marks) |
**Part (b)(i)**
$\sqrt[3]{8}\!\left(1 + \frac{3}{8}x\right)^{\frac{1}{3}}$ | B1 | $8^{\frac{1}{3}}(1+kx)^{\frac{1}{3}}$
$= 2\!\left(1 + \frac{1}{3}\!\left(\frac{3}{8}x\right) - \frac{1}{9}\!\left(\frac{3}{8}x\right)^2\right)$ | M1 | Replacing $x$ with $kx$ in answer to (a)
$= 2 + \frac{1}{4}x - \frac{1}{32}x^2$ | A1 (3 marks) | For numerical expression which would evaluate to answer given
Alternative: B1 – all powers of 8 correct: $8^{\frac{1}{3}}\ 8^{-\frac{2}{3}}\ 8^{-\frac{5}{3}}$; M1 – powers of $3x$ (condone $3x^2$); $2 + \frac{1}{2} \cdot \frac{x}{8^{\frac{1}{3}}} - \frac{1}{9} \cdot \frac{1}{8^{\frac{5}{3}}} \cdot 9x^2$; A1 – see some arithmetic processing, must see 9s in last term
**Part (b)(ii)**
$x = \frac{1}{3}$: $\sqrt[3]{8+1} = 2 + \frac{1}{4}\times\frac{1}{3} - \frac{1}{32}\times\left(\frac{1}{3}\right)^2$ | M1 | Using $x = \frac{1}{3}$ in given answer
$\sqrt[3]{9} = \frac{576+24-1}{288} = \frac{599}{288}$ | A1 (2 marks) | Any correct numerical expression $= \frac{599}{288}$
**Total: 7 marks**5
\begin{enumerate}[label=(\alph*)]
\item Find the binomial expansion of $( 1 + x ) ^ { \frac { 1 } { 3 } }$ up to the term in $x ^ { 2 }$.
\item \begin{enumerate}[label=(\roman*)]
\item Show that $( 8 + 3 x ) ^ { \frac { 1 } { 3 } } \approx 2 + \frac { 1 } { 4 } x - \frac { 1 } { 32 } x ^ { 2 }$ for small values of $x$.
\item Hence show that $\sqrt [ 3 ] { 9 } \approx \frac { 599 } { 288 }$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C4 2007 Q5 [7]}}